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draw-position-time-graph-for-constant-acceleration

๐Ÿš€ Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In uniformly accelerated motion, an object moves with a constant acceleration, meaning its velocity changes at a constant rate. This concept is crucial for understanding how objects move in a straight line under the influence of constant acceleration. The position-time graph for an object undergoing constant acceleration is a parabola, indicating that the position changes non-linearly with time. The equations of motion can be used to describe the relationship between displacement, initial velocity, time, and acceleration.

Theory Explanation

Understanding Uniformly Accelerated Motion

In uniformly accelerated motion, the acceleration (a) is constant. This means that the change in velocity (ฮ”v) over a time interval (ฮ”t) is the same throughout the motion. The basic equations of motion can be derived from this concept, which relate displacement, initial velocity, final velocity, acceleration, and time.

\[ v = u + at, \ s = ut + \frac{1}{2}at^2, \ v^2 = u^2 + 2as \]
Position-Time Graph for Constant Acceleration

When plotting a position-time graph for an object in uniformly accelerated motion, the graph will be a parabola. The shape of the graph indicates that the position of the object changes at a rate that increases over time due to the constant acceleration. The initial position can be represented as the y-intercept, and the curvature of the graph represents the acceleration.

\[ s(t) = s_0 + ut + \frac{1}{2}at^2 \]

Key Points

  • ๐ŸŽฏ Uniformly accelerated motion involves constant acceleration.
  • ๐ŸŽฏ The position-time graph for constant acceleration is a parabola.
  • ๐ŸŽฏ The equations of motion can be used to calculate displacement, velocity, and time.
  • ๐ŸŽฏ Initial velocity (u) and acceleration (a) are key parameters in these equations.
  • ๐ŸŽฏ Understanding the shape of the graph helps in visualizing the motion.

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Examples:💡

A car starts from rest and accelerates uniformly at 2 m/sยฒ for 5 seconds. Find the distance covered by the car during this time.

Solution:

Step 1: Given: Initial velocity (u) = 0 m/s, acceleration (a) = 2 m/sยฒ, time (t) = 5 s. We will use the equation s = ut + 1/2 atยฒ.

\[ s = 0 \cdot 5 + \frac{1}{2} \cdot 2 \cdot 5^2 \]

Step 2: Calculating the distance: s = 0 + 0.5 * 2 * 25 = 25 m.

\[ s = 25 m \]

An object is thrown upwards with an initial velocity of 20 m/s. Calculate the maximum height reached if the acceleration due to gravity is -9.8 m/sยฒ.

Solution:

Step 1: Using the equation vยฒ = uยฒ + 2as, where final velocity (v) at the maximum height is 0 m/s, initial velocity (u) = 20 m/s, and acceleration (a) = -9.8 m/sยฒ.

\[ 0 = 20^2 + 2(-9.8)s \]

Step 2: Rearranging gives: 0 = 400 - 19.6s, thus 19.6s = 400, so s = 400/19.6 = 20.41 m.

\[ s = 20.41 m \]

Common Mistakes

  • Mistake: Confusing the signs of acceleration when dealing with upward and downward motion.

    Correction: Always remember that acceleration due to gravity is negative when an object is moving upwards.

  • Mistake: Forgetting to convert units when necessary, especially in mixed unit problems.

    Correction: Always check that all units are consistent before performing calculations.

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